help-27

[torn moat]

teacher and student do work 3.5 times faster than just the student how much faster can the teacher do the work compared to student

[torn moat]

<@&286206848099549185>

<@&286206848099549185>

nvm

[tiny iris]

Let's say the student does the work for time x and the teacher for time y.
x + y = 3.5x
y = 3.5x - x
y = 2.5x
So 2.5 times faster

[devout snow]

@torn moat Has your question been resolved?

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[restive river]

The other channel closed

And no one responded

[slow mist]

no it didn't

i did

[restive river]

What

Nvm I see it

Can you close this one then for me

[slow mist]

go back to #help-17

just type .close

haaa well fell the same

[restive river]

Ok

I’m confused

[slow mist]

it's fine i was confused when i joined first too

but don't rush to get answers

[restive river]

I know the answer

I’m just wondering better what to get it

[slow mist]

sometimes it might take an hour to get responded

[restive river]

Way

Ok

[slow mist]

what was your way ?

[restive river]

It was just graphing |x^2-4| -2x and then seeing where the horizontal line k would intersect 3 points

It’s not a good way to do it because it’s like an approximation

[slow mist]

and what was the answer ?

[restive river]

I saw it was when k=4 or k=5

[slow mist]

you say it's a bad way but it's actually a very good way to find this kind of problems

[restive river]

Is there some kind other kind of way though

[slow mist]

if it wasn't the only

[restive river]

My graph was different cuz I did a different way

But same thing

You have any idea how to do it without looking at graph?

[slow mist]

if you ask me this is the best way

[restive river]

Well imagine you didn’t have a graphing website or calculator, all you would have is a rough estimate then if you did the graph like me on a paper

It’s like solving a systems of equations by graphing, not exactly accurate

[scarlet sequoia]

consider two cases
first (-inf, -2] v [2, inf) ---> f(x) = x^2 - 4
second (-2,2) ---> f(x) = 4 - x^2
now first case is:
x^2 - 4 = 2x + k ---> x^2 - 2x - k - 4 = 0
second one:
4 - x^2 = 2x + k ---> x^2 + 2x + k - 4 = 0
now first equation can have two solutions and the other one can have one solution or the other way around

[restive river]

I did that

But discriminant weren’t 4 in one case and 5 in other

One was 5 but it didn’t give me 4

For the other

[scarlet sequoia]

D1 = 4(k+5)

[restive river]

Yes

[scarlet sequoia]

D2 = -4(k-5)

[restive river]

Yes

[scarlet sequoia]

now D1 > 0 and D2 = 0

[restive river]

k=5

That one works

[scarlet sequoia]

gives k > -5 and k = 5 which is finally k = 5

[restive river]

but k=4, how?

Other cases gives k=-5

Case*

That is how I did it too but it didn’t work

[scarlet sequoia]

because there is one more case

D1 > 0 and D2 > 0 and one solution is of multiplicity 2

aka it reapeats

[restive river]

Oh

So how would that work then

That system gives k>5, no?

Nevermind

I messed up

[scarlet sequoia]

it gives -5 < k < 5, but there is one more restriction

[restive river]

k<5

yes

What is the that restriction

?

[scarlet sequoia]

mhm I'd say you can find solutions in terms of 'k' and then equate them in order to solve for k

but remember about the domain of x

[restive river]

-2<x<2 when inside of modulus is negative and x<=-2 or 2<=x when inside of modulus is positive, right?

How would I know which solutions to equate

Wait is it just when discriminant is 0?

Nvm that’s just an identity

[scarlet sequoia]

ok other way

faster

you have two parabolas now

equate them

find x in terms of 'k'

you'll get x = -k/2

substitute x into their equations then solve for k (remember about domain of x !!!)

you'll get k = 4

[restive river]

Thank you

[scarlet sequoia]

hehe but you see graphing is more efficient for questions like this

[restive river]

.close

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[cursive totem]

Suppose $f'(x)>0$ on some interval $(a,b)$ so that $f(x)$ is strictly increasing. Show that the inverse is differentiable on $(a,b)$.

[woven radish]

please request a new nickname

[cursive totem]

I tried using the derivative definition to get...

let g(x) be the inverse

$\lim_{t\to x} \frac{g(t) - g(x)}{t-x}$

[woven radish]

please request a new nickname

[cursive totem]

Then, since g is just the inverse there are some real numbers r,s such that ...

$\lim_{f(r)\to f(s)} \frac{r -s }{f(r) - f(s)}$

i.e., f(r) = t and f(s) = x

im thinking that this is just the inverse of f'(x)

1/f'(x)

but im not sure this really makes sense

[woven radish]

please request a new nickname

[cursive totem]

$=\frac{1}{f'(s)}$

[woven radish]

please request a new nickname

[cursive totem]

like that's just nonsense right?

[vast rain]

well, I haven't looked through your proof

but the result is supposed to be something like that

where

$\dv{f^{-1}}{x} = \frac{1}{f'(f^{-1}(x))}$

[woven radish]

Saccharine

[devout snow]

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[devout snow]

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[vital edge]

I don't understand why it doesn't work
def P(x):
return 2x3-3x2-1

def encadrement():
x=1
while P(x)<0:
x=x+0.1
return x-0.1,x

[pseudo basin]

no goal was given.

[vital edge]

so the goal that I have to establish will come from where?

[topaz axle]

2x**3 - 3x**2 - 1 is not syntactically valid

[vital edge]

ah ok thank you so how can i write this ?

[topaz axle]

2*x**3 - 3*x**2 - 1

[vital edge]

thank you

[devout snow]

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[radiant drift]

hi u compute it like the 2nd pic right?

[pseudo basin]

why ping me like this

but yes that's one way of computing it

i would have preferred 11!/(5! 2! 1! 1! 2!) though

[devout snow]

@radiant drift Has your question been resolved?

[devout snow]

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[devout snow]

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[scenic surge]

Stupid question but like whats the point of FTC1?

[scenic surge]

Like I dont get it

[lusty sapphire]

Ftc1?

[scenic surge]

I get the importance of FTC2 But not 1

fundamental theorem calc 1

one sec ill send poc

pic

top right

[violet wind]

well it helps you solve this derivative for one

[scenic surge]

whos derivative

F(x)?

[violet wind]

yeah

[lusty sapphire]

Therefore the derivative of the definite integral

[violet wind]

FTC1 is basically derivatives and integrals are opposite operations

maybe it's not usually referenced directly but it's called fundamental for a reason, important stuff is kind of build up on it

[lusty sapphire]

Some folks take this fact for granted, forgetting that definite integral is defined as a riemann sum and not antiderivative

[violet wind]

integration by parts is the opposite of product rule, and u-substitution is the opposite of the chain rule

[violet wind]

^ yeah this is really important

derivatives and integrals each have their own independent meaning, derivatives being about rate of change and integrals being about a kind of sum

it's entirely possible to talk about what one is like without ever mentioning the other

[scenic surge]

I see

when I was readin it at first I thought it was kinda pointless

[violet wind]

ah yeah

that it's not true by definition is probably the important part

[scenic surge]

FTC2 was a game changer doe lol

for me

ty

[violet wind]

ah yeah

its basically what lets you evaluate definite integrals

so makes sense

np

[violet wind]

^without an incredible amount of pain

it would still be possible otherwise, you'd need to use the riemann sum definition

[scenic surge]

Mhm so its literally just plug in the upper bound inside f(t) and times it by derivative of the upper bound

always for FTC1 ^

[bright burrow]

What is ftc2

nvm googled it💀

[scenic surge]

mhm

.close

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@proper thicket Has your question been resolved?

[gloomy prairie]

How do you calculate probability？

[lusty sapphire]

Factor out sqrt(2) where you can

[devout snow]

@proper thicket Has your question been resolved?

[lunar harbor]

.close

[devout snow]

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[devout snow]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[tame sage]

2

[lunar harbor]

Send what you have then 🤷‍♂️

[tame sage]

.close

[devout snow]

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[amber mural]

im confused what it wants

i just did summation k=3 to ... of the probability function

wait should i cap summation at 1 and solve for upper lim?

[lunar harbor]

If I had to guess, they want $\sum^{\infty}_{k=3} P(k)$, but the question could definitely be worded better.

[woven radish]

Civil Service Pigeon

[amber mural]

ok but heres my question

how can u find the sum to infinity

doesnt it need to be finite or something

[lunar harbor]

It is finite

Note that you can use the formula $\sum^{\infty}_{k=0} ar^k=\frac{a}{1-r}$ if $|r|<1$

[amber mural]

oh

[woven radish]

Civil Service Pigeon

[amber mural]

ok thanks

[low holly]

1 - P(0 or 1 or 2)

[amber mural]

ye that works

[lunar harbor]

ooh good point 👀

[amber mural]

ill write that too

[lunar harbor]

Wait isn't it 1-P(0, 1, or 2)

[amber mural]

ye

bruh why do i need calc 3 as stats major that thing is killing me

[devout snow]

@amber mural Has your question been resolved?

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[opal owl]

<@&286206848099549185>

[devout snow]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[opal owl]

ok

Helllo?

[grizzled yew]

Where is the question

[opal owl]

wdym

[torn wave]

did you read #❓how-to-get-help

[opal owl]

yes

[torn wave]

so ask your question please

[upper schooner]

…and what would you like help with, that you opened the channel for?

[opal owl]

unit rates

[torn wave]

go on

[opal owl]

I Never knew how to do it because our math teacher cured at us and left

.close

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[gleaming wedge]

hi

could someone help me

[upper schooner]

You have #help-29

.close

[devout snow]

Channel closed

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Use .reopen if this was a mistake.

[upper schooner]

.reopen

[devout snow]

✅

[upper schooner]

Oh you just closed it nvm

[gleaming wedge]

oh

ok

can you help me?

[upper schooner]

I'm not too good with these types of problems sorry

[gleaming wedge]

oof ok

<@&286206848099549185>

[upper schooner]

Someone will come and help though hopefully if you wait

[gleaming wedge]

oki

<@&286206848099549185>

<@&286206848099549185>

can anyone help me

;-;

[wooden wraith]

You have a few right triangles in there

It's probably a good idea to give names to some of these segments so we can work with them easier

particularly QT, TR, RS, and SP

maybe start by labeling those a, b, c, and d

[gleaming wedge]

ok

[wooden wraith]

then use the pythagorean theorem to see what kind of relationships you can come up with

[gleaming wedge]

so now what

but how

[wooden wraith]

so like this

[gleaming wedge]

ok

[wooden wraith]

Pick a right triangle and use the pythag theorem, let's just see what happens

[twilit ether]

Wouldn't... ST/QS= PT/PQ?

[gleaming wedge]

how though

[wooden wraith]

do you know the pythagorean theorem?

[gleaming wedge]

yes

[wooden wraith]

so like triangle SRT

what does the pythag theorem say about SRT?

[wooden wraith]

🤔 I don't think so

[gleaming wedge]

we know that

b^2+c^2=4

[wooden wraith]

perfect awesome

What about QRS?

[gleaming wedge]

hh

c^2+(a+b)^2=16

[wooden wraith]

yes perfect

and what about PRT?

[gleaming wedge]

b^2+(c+d)^2=25

[wooden wraith]

okay great

And since we want to find PQ, do you also agree that

(a+b)^2 + (c+d)^2 = (PQ)^2

[gleaming wedge]

yes

[wooden wraith]

So if we could find a value for (a+b)^2 + (c+d)^2, then we could find PQ

[gleaming wedge]

yes

[wooden wraith]

so I'm gonna copy/paste your three equations from before

b^2+c^2=4
c^2+(a+b)^2=16
b^2+(c+d)^2=25

Any way we can combine those to get (a+b)^2 + (c+d)^2 ?

[gleaming wedge]

yes

1 second

ill be back in a second

[wooden wraith]

ok

[gleaming wedge]

ok sorry

im back

?

[wooden wraith]

okay

[wooden wraith]

so do you see how we can combine the equations to get what we want?

[gleaming wedge]

ye

so like

c^2=b^2+9?

or c = b+3?

[wooden wraith]

hm, not quite. You can't split up a square root over addition like that

[gleaming wedge]

oh ok

[wooden wraith]

and I'm not sure exactly where this came from

[gleaming wedge]

nvm on that

[wooden wraith]

What do you get if you add the second and third equations together here

[gleaming wedge]

uhh

idk

41

?

[wooden wraith]

on the right side, 41 yes

what about on the left?

[gleaming wedge]

b^2+c^2+(a+b)^2+(c+d)^2

[wooden wraith]

Yes awesome

b^2+c^2+(a+b)^2+(c+d)^2 = 41

And those first two terms, b^2 + c^2

We know what that equals

[gleaming wedge]

4

[wooden wraith]

ok, so..

[gleaming wedge]

sqrt37 is my answer

[wooden wraith]

Bingo 👍

[devout snow]

@gleaming wedge Has your question been resolved?

[devout snow]

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[plucky glacier]

why isn't 1 a vertical asymtote?

[pseudo basin]

factor the numerator too

you should get that (x-1) is a factor of it as well

so you get a hole instead

[plucky glacier]

oh so since the numerator equals 1 when set equal to 0, its a hole instead of an asymtote?

[pseudo basin]

bad phrasing

the numerator factors as (x-1)(2x+1) by the looks of it

[plucky glacier]

ah ok thxx

.close

[devout snow]

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[plucky glacier]

Idk what Im doing wrong T-T

[upper schooner]

Bear in mind the sections where the function is defined

Similarly

[plucky glacier]

I still dont get it

[upper schooner]

You put in 3 as a value, but it doesn't count because of the restriction here, that x<1

[upper schooner]

For similar reasons, -1/2 doesn't count either

Note your function looks like this basically https://www.desmos.com/calculator/nbfiycujcj

[upper schooner]

at x=3 you're "a whole new function" (and similar for x=-1/2) if that makes it any better?

[plucky glacier]

ok thxx

.close

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[fervent jewel]

Does this make sense?

[fervent jewel]

.close

[devout snow]

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[eager crystal]

the only answer i can come up with this problem is that value

5pi/4

but the answer key says -pi/4

what could i possibly be doing wrong

[sick fulcrum]

sin inverse sin doesn't exactly cancel like this

[winter patrol]

you didn't consider the range of the arcsin function

[eager crystal]

aw the ranges always mess me up

i know the range for arctan is from -90 to +90

let me look up the range for arcsin

looks like −π2,π2

so -360 to +360?

isn't it the whole circle then

[upper schooner]

Why hey there again Snowball 🙋‍♀️

[sick fulcrum]

,w plot arcsin(x)

[eager crystal]

hello

[eager crystal]

.

[upper schooner]

$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

[woven radish]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by DerpZ#9141

[eager crystal]

oooh

that 2 means over 2?

[sick fulcrum]

chartbit typoed but

its -pi/2 to pi/2 lol

[woven radish]

chartbit

[sick fulcrum]

yup thats what they meant

[eager crystal]

how is this over 2

[sick fulcrum]

😩

[eager crystal]

weird notation

im really confused

[sick fulcrum]

google smh

yeah that's not corrcet

correcr

[upper schooner]

Poor typesetting

[sick fulcrum]

,w range of arcsin function

[woven radish]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by DerpZ#9141

[sick fulcrum]

bad wolfie

[eager crystal]

aw i see thanks

so

[eager crystal]

it would be -90 to +90

isn't it the same as arctan?

first quad and 4th quad?

that reminds me of arctan

[upper schooner]

Yep arctan gives you a value between -90 and 90

But in the case of arctan it excludes -90 and 90

But with arcsin it includes both

[eager crystal]

oh i see

so going back to the problem

[upper schooner]

[I remember helping you with these types of questions some time ago ]

[eager crystal]

the answer would be

in the range

so since 5pi/4 aka 225 degrees is not in range we gotta find something in the range that gives the same value

so is it pi/4?

this is still not what the answer key says

[sick fulcrum]

close but not quite

recall your quadrants

[eager crystal]

pi/4 is in the first quadrant

[sick fulcrum]

is 5pi/4 in the first quadrant?

[eager crystal]

no

2nd

ooooh

so the opposite would be

4th quadrant

so the answer is in the 4th quadrant

which is -pi/4?

oh that makes sense

much appreciated guys

🙂

.close

[devout snow]

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[eager crystal]

oh just one more quick question

[devout snow]

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[eager crystal]

do you have the ranges of arctan, arcsin, arccos memorized or is there a way to derive to them?

[sick fulcrum]

you know range of sin

cos and tan

and the inverse trig functions have the same range as the domain of sin cos and tan

well

the domain where they are 1-1 and onto

[upper schooner]

For me it's pretty much a memorise thing tbh

[sick fulcrum]

true

or if you have access to a graphing calculator

use that

[eager crystal]

like is it just mathematicians decided to set the ranges or is there's a logical reason behind all the ranges?

[sick fulcrum]

well

[sick fulcrum]

^^

[vast rain]

part of both

basically, it can't be all real numbers because then you'll have trouble actually defining them properly

in the sense that multiple answers come out

[eager crystal]

ooh the ranges are the same as the domains of sin, cos , and tan?

[vast rain]

but also the specific intervals over which they're defined are kinda arbitraryish

[sick fulcrum]

not quite

not the full domain

but we restrict it so that those functions are

well

[eager crystal]

yeah exactly

[sick fulcrum]

,w graph sinx from -pi/2 to pi/2

[eager crystal]

i also think it's half and half

[woven radish]

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[sick fulcrum]

we restrict the domain of sinx here to ensure only 1 thing comes out

of course we could've restricted it to some other range

like say

[eager crystal]

oooooooh

[sick fulcrum]

,w plot sinx from 3pi/2 to 5pi/2

[woven radish]

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[sick fulcrum]

like this for example

but -pi/2 to pi/2 is nice

and less annoying

chartbit on point with the reacts

[eager crystal]

oh that makes sense

[sick fulcrum]

if you want to sound fancy

[eager crystal]

you can prove it but you also gotta memorize part of it right?

[sick fulcrum]

we call that the restriction of sin to -pi/2 to pi/2

[eager crystal]

so it's like half and half

[sick fulcrum]

well

yeah

I mean its probably easier to memorize it

,w plot tanx on (-pi/2, pi/2)

oh useless

[woven radish]

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[sick fulcrum]

sigh

but you get the point

[upper schooner]

,w tan(x) from -pi/2 to pi/2

[woven radish]

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[eager crystal]

now i want to prove the restriction of cos since the range of arccos is the only one that's different from the other 2

[sick fulcrum]

[upper schooner]

Alright I give up

[sick fulcrum]

wait hold on lemme try smth

,w y=tan(x), y=0, y=0 from -pi/2 to pi/2

[woven radish]

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[sick fulcrum]

yeah so for cos its slightly different

[eager crystal]

so i need to restrict the domain of cos so that it only has one value

[sick fulcrum]

well it only maps to one value yes

restrict that domain

a "fun" exercise is to pick different domains and see if you can redefine arcsin/arccos etc

[eager crystal]

so i need a function first

lets say it is y=cos(x)?

cos(pi) would give me -1

so i need to stop at where it gives me -1 again

that's how i restrict right?

so just adding 2pi?

which is 3pi

[sick fulcrum]

not quite

if you do that

youll ge

,w plot cosx from pi to 3pi

[woven radish]

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[eager crystal]

umm

[sick fulcrum]

which is not 1-1 and onto

or here

[eager crystal]

what is 1-1 and onto?

[sick fulcrum]

it gives the same value for multiple different x

[eager crystal]

im not following lol

[sick fulcrum]

ok for now you can think of it as

see how for

2 different values of x

sin(x) can be the same?

we don't want that

we want sin(x) to be different for different values of x

[eager crystal]

yes

ooh

i see

[sick fulcrum]

so try restricting again

[eager crystal]

ye there's only one x intercept

but mine has 2

alright lemme try again

[eager crystal]

im first looking at this to find what are the x values that are intercepting the x axis

looks like one is right before 5

like 4.8

the other one is like 7.8

what are those two numbers coming from?

[upper schooner]

,calc 3*pi/2

[woven radish]

Result:

4.7123889803847

[eager crystal]

ooooh

[trim vine]

For me, I find them somewhat not arbitrary. When you think of sin, you think of the vertical y axis of the unit circle, going from y= -1 to 1. If you try think of a section of the circle that has points that give y values from -1 to 1, the first set points that comes to mind is the ones with angle -pi/2 to pi/2. If you think of the region of the circle between -pi/2 to pi/2, the y-axis makes up a wall of that region. And as for the horizontal x axis, the section of the circle that has points ranging from x= -1 to 1, the first set that comes to mind if 0 to pi. That region has the x-axis as its wall. I also notice both of these share the first quadrant.

[eager crystal]

i see now

[eager crystal]

thanks

3pi/2

[upper schooner]

It's similar for the other one, 5pi/2

[eager crystal]

yes

so i would have to restrict so that it doesn't contain 5pi/2

[eager crystal]

this one is to 6pi/2 aka 3pi

so what if I

[upper schooner]

Basically you would need to restrict it such that you go from -1 to 1 (or 1 to -1)

[eager crystal]

oh by -1 and 1 are you talking about the y values?

[upper schooner]

That way you get all the values you need, but then go more than that and you'll end up repeating yourself, if thats anywhat clear?

[eager crystal]

i see

[upper schooner]

Yeah

Like you could start from pi and go to 2pi for example

Or you could start from 2pi and go to 3pi

[eager crystal]

so on the unit circle 0 or 2pi would give me 1

and pi would give me -1

so now i have both 1 and -1

so is the restricted domain for cos from 2pi to pi?

or from pi to 2pi?

if the order matters

[eager crystal]

seems like this is fine

just googled it

nice now all i have to remember is 1 and -1

i will not forget the range of arc (tan, cos, sin) for the rest of my life

thanks guys

!

[upper schooner]

The order kinda matters in a way if you're writing it as an interval, but if you're just saying it we can figure out what you mean

Basically you would need to write [pi, 2pi] as the interval you're choosing, but otherwise if you just say where you're considering then most people will get you

[eager crystal]

so is it closewise?

if you say [pi,2pi]

[upper schooner]

Hmm what do you mean by closewise? It is a closed interval yea, because you need to include both pi and 2pi

[eager crystal]

like on the unit circle

if you give me the unit circle and say [pi,2pi], i would have to know whether it's clockwise or counter clockwise

i might be thinking of 3rd and 4th quad while you are thinking of 1st and 2nd

that's why i was asking

looks like it is clockwise

after some research

[eager crystal]

wait then this is wrong

[pi, 2pi] is counterclockwise

shouldn't we say [0, pi] instead?

[upper schooner]

Hmmm, hang on, do you need them to be (counter/)clockwise?

[eager crystal]

i think it is just clockwise

but this [pi, 2pi]

is wrong if you follow clockwise

it would give you 3rd and 4th quad

[upper schooner]

Hmmm, just to clarify, this is what you're trying yea?

[eager crystal]

but we want first and 2nd

[eager crystal]

yes

i might be just tripping

so what im asking is that

is [pi, 2pi] the same as [0, pi]?

if you count clockwise each of them give you different quadrants

[upper schooner]

Wellll, in terms of being able to define an inverse they are

[eager crystal]

oh i see

[eager crystal]

so this is the visual of what i was talking about

but it seems like both are fine

in terms of defining inverses

like those two images are different

so i was wondering

both are clockwise

[upper schooner]

Ahh I get you a bit more now, yea in any case because they're intervals you'd be going anticlockwise

[upper schooner]

They're anticlockwise btw

[eager crystal]

wait

wow

im dumb

yes

this whole time

loool

yeah i meant anticlockwise

[upper schooner]

You start from a lower value then go to a higher one

[upper schooner]

Don't worry, it happens

[eager crystal]

but if you go anticlockwise from pi to 2pi it gives you 3rd and 4th quadrants

well thank you so much for spending time with me on those trig concepts :happyCat:

opps

i can't do the emoji

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[devout snow]

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[soft ferry]

Hello, Im an grade 9 student and I like to do random math questions for fun. Right now Im stuck on part B. I solved part A and got 0.27-im pretty sure its right , but for question B I simply don't know how to solve it. Does the answer have something to do with the velocity ratio from part A?

[lusty sapphire]

Yes

[soft ferry]

I searched the question up and apparently the answer for part B is 4.5m/s , but the explanation wasn't provided. I've tried and can't reach the same answer.

I think an explanation on why the answer is 4.5m/s would be helpful

<@&286206848099549185>

[devout snow]

@soft ferry Has your question been resolved?

[coral tulip]

Your VR= 4/15

$V.R. =\frac{V of effort}{V of load}$

[woven radish]

Anagh

[coral tulip]

V is velocity

@soft ferry

[soft ferry]

yeaa and i got 0.27 for the velocity ratio

would that be correct?

[coral tulip]

Don't convert it into decimal without getting final answer

It will be easier for calculation

[coral tulip]

0.27 is the round off

[soft ferry]

Ohh

I see why my answer was wrong

I rounded first

[coral tulip]

What was your answer?

[soft ferry]

16200m/h

ohh so it was 4.5m/s

[coral tulip]

Not exactly

But round off

[soft ferry]

um so would the exact answer be 40/9 ?

I've got it now

Thanks for your help

.close?

.close

[devout snow]

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[still jacinth]

what does it mean to

[restive river]

Trig?

[hushed wraith]

can u send the entire problem?

[still jacinth]

i got around 3.3084

[restive river]

I assume they mean by a full time period which would be to the next half year?

[still jacinth]

ohh thank u!

oh that makes sense

like period of time

tysm!

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[fiery yoke]

Hello Guys!
Quick question regarding the root test for a series. Does the test work for any power (k-2,k+2, 2-1k) etc. or is there a specific requirement to cancel out the root?

[hybrid snow]

Wdym any root

Like the base of the root?

[fiery yoke]

Yes

[hybrid snow]

It has to be n

Have you learned limit superiors yet

[fiery yoke]

Wrong formulation, I meant for any power of the series we are going to apply the root test

[fiery yoke]

No not yet

[hybrid snow]

Just worry about $\limit{\sqrt[n]{|a_n|}}{n}{\infty}$

[woven radish]

Umbraleviathan

[fiery yoke]

But my n can be wahtever?

2k, k, 2, 2-k?

[hybrid snow]

Well in your case, n = k

Typically n is used for indices

[fiery yoke]

Yes, but does it matter what specific n or k it is?

[hybrid snow]

No it shouldn't, not ultimately. But it has to correspond to the index

So you can't let n = 2k, etc etc

Because typically summations use n instead of k

For your purpose, your root test (matched up with your index variable) would be $\limit{\sqrt[k]{|a_k|}}{k}{\infty}$

[woven radish]

Umbraleviathan

[fiery yoke]

Yes, thanks!

I just wanted to check if I need to be considered about any special cases

[hybrid snow]

Ngl n is a dummy variable

[fiery yoke]

But I think for the basic level at the moment it should be fine

[hybrid snow]

It could be j, etc

[fiery yoke]

Yeah ofc

[hybrid snow]

It's just whatever the index variable is

[fiery yoke]

But it is whatever index

[devout snow]

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[fiery yoke]

.close

[devout snow]

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[frosty gyro]

Is triangle fed similar to cba

[devout snow]

@frosty gyro Has your question been resolved?

[knotty bolt]

its hard to see, but if FED is the triangle that connects all the middle points of ABC then it is

[frosty gyro]

De and Fd are parallel to ab and ac

[knotty bolt]

looks bad

[frosty gyro]

Wdym

[knotty bolt]

its annoying, you likely need a bigger drawing so u can fit all angles

DE||AB means that angle ABC = angle EDC

same for mirror

FD||AC means that angle BCA = angle FDB

use that to find angle FDE

[frosty gyro]

ok

[frosty gyro]

angle FDE= angle A?

[knotty bolt]

yes

[frosty gyro]

so it is similar

also I wrote that already

[knotty bolt]

so it would seem

[frosty gyro]

in the pic

[knotty bolt]

I have bad eyes

[frosty gyro]

my bad

then bdf is similar to cba

and dec is similar to bac

and aef is similar to acb

fe is parallel to bc

if fe is parallel to bc, bfd and dec have the same height, so bd=25dc?

but if they're similar, bd=5dc

so idk which

[devout snow]

@frosty gyro Has your question been resolved?

[frosty gyro]

@dry robin

[frosty gyro]

.close

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[worthy berry]

im stuck on part c

[worthy berry]

i've got it into cos(cos^-1(sqrt(2)/5)) = cos(x-tan^-1(11/2))

then tanx=tan(tan^-1(11/2)+cos^-1(sqrt(2)/5))

or tanx = tan(tan^-1(11/2)-cos^-1(sqrt(2)/5))

but then the answers say im wrong

OMG I GOT IT

it's not sqrt(2)/5

it's 2/sqrt(5)

🤦

thanks for the help guys

i finally finished the question

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[azure widget]

Hello. I apologise if I am interrupting your time.

[azure widget]

I have a unclear exercise.

Line AB, where A(-3,-2), B(9, 10) divide: B) on three equal piece C) on three parts where relations are 1:2:3.

B) and C) are unclear.

If someone could help, I'd appreciate it.

In advance thank you very much.

[torn vessel]

For B, find two points on the line so you create 3 segments that are all the same length

[azure widget]

What are segments?

[devout snow]

@azure widget Has your question been resolved?

[devout snow]

@azure widget Has your question been resolved?

[devout snow]

@azure widget Has your question been resolved?

[azure widget]

No one did solved my problem : (.

[devout snow]

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[lucid falcon]

Hello, i need help with this problem

[lucid falcon]

Anytime i run into square roots, my brain goes blank

[cursive jacinth]

Do you know what it means when a line is tangent to the graph

[lucid falcon]

im trying to get a line from one point on the line to another right?

[cursive jacinth]

Yes

And that line is tangent to the graph g(x)

[hybrid snow]

,tex \linearization

[woven radish]

Umbraleviathan

[hybrid snow]

This is the equation for a tangent line of a function at x = x_0

Derived from point-slope form

[lucid falcon]

okay okay so how would we find our x of 0

[hybrid snow]

It's quite literally given to you

[lucid falcon]

or is that the provided point

[hybrid snow]

_ _

[cursive jacinth]

Do you know what it means when a line is tangent to a graph?

[hybrid snow]

It says that youre finding the tangent line at x = 4

so x_0 is 4

[lucid falcon]

ah okay okay

so then i need to find f(4) to plug into the equation for a tangent line

it there anyway to do voice chat with this or no haha

[cursive jacinth]

If a line is tangent to a graph it means the line has the same derivative as the graph

[lucid falcon]

the problem provides 2 equations how do i know which one to use for what

[hybrid snow]

Okay

I really hope you can read

[hybrid snow]

I gave you this

[cursive jacinth]

You calculate the derivative first

[hybrid snow]

It gives you the derivative already

[lucid falcon]

ill show what i wrote down

[hybrid snow]

The only calculations you're gonna need to do is g(4) and g'(4)

But g(x) and g'(x) are already given to you

,tex \linearization

[woven radish]

Umbraleviathan

[hybrid snow]

And then I gave you this

[lucid falcon]

im sorry im not the best at this but im trying to comprehend everything

[hybrid snow]

x-4, not x - 1/2sqrt(4)

[hybrid snow]

Look up here

It explicitly says "(x-x_0)"

[lucid falcon]

okay okay

so then it was the one i wrote on top

[hybrid snow]

And also if you understand point slope form it should've made no sense for it to be (x-1/2sqrt(4))

Everything else is fine

[lucid falcon]

i wrote the one on top first thinking it was that

[hybrid snow]

I'm more concerned about your understanding of tangency. You're familiar with point-slope form for lines right?

[lucid falcon]

i understand point slope and how getting tangent lines its just getting point slope with extra steps

[hybrid snow]

Essentially point-slope form will be your best friend for tangent lines

You can think of (x_0, f(x_0)) being the point the tangent line passes through

and f'(x_0) being the slope of that line

[lucid falcon]

okay okay

im going to try it out and post what i got

i was just over complicating things

.close

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[brazen agate]

https://gyazo.com/33169df14a94e10f4f43f20df583e78e

[brazen agate]

i just started to study this but how would i find the elementary row operation

i know that they top row changed but idk what they did

[hybrid snow]

It legit just looks like R1 + 5(R2)

[brazen agate]

oh

wait

lol

thank you

/close

.close

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[cursive totem]

How many 3-letter anagrams are there of ARKHAMIM

[cursive totem]

My answer was ...

8x7x6/(2!2!)

where i divide by 2!2! because there are two A's and two M's

but I was told this was wrong

[fathom thicket]

what was wrong, your final answer?

[cursive totem]

this is exactly how I submitted it but apparently im wrong?

[fathom thicket]

you submitted this?

and not an actual number?

[cursive totem]

yeah

[fathom thicket]

are you supposed to do that?

[cursive totem]

yeah

they want to see where my work applies

.close

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[graceful cedar]

if A, B, C, D are any sets and ... then |A [reverse u] B|...

[devout snow]

Please don't occupy multiple help channels.

[graceful cedar]

how can i prove it?

[devout snow]

@graceful cedar Has your question been resolved?

[graceful cedar]

<@&286206848099549185>

[cursive totem]

Do you know the general formula for |AUB|

In general for two sets:

$|A\cup B| = |A| + |B| - |A\cap B|$

[woven radish]

please request a new nickname

[cursive totem]

We have

$|C\cup D| = |C| + |D| - |C\cap D|$

[woven radish]

please request a new nickname

[cursive totem]

but these intersections are empty

so its really just comparing...

$|A| + |B|$ and $|C| + |D|$

[woven radish]

please request a new nickname

[devout snow]

@graceful cedar Has your question been resolved?

[devout snow]

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[devout snow]

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[frozen citrus]

I need help for 2 and 3

[wicked turtle]

what methods are you allowed to use?

do you know about taylor series for example

[dim shadow]

cant u use l'hospitals for 2

[wicked turtle]

actually a trig identity makes it easy

[frozen citrus]

I dont know how to do that

[wicked turtle]

do you know one for cos(2x) ?

[frozen citrus]

hm?

[wicked turtle]

cos(2x) = cos^2(x) - sin^2(x), does that sound familiar

[frozen citrus]

....no?

[inner pilot]

here is the solution of 2

[wicked turtle]

i bet you knew it once 😁

[dim shadow]

,rotate

[frozen citrus]

thank u

[wicked turtle]

what's the point of just writing out an answer for the OP

[inner pilot]

wait

[frozen citrus]

how do I do l'hospital?

[inner pilot]

here it is

[frozen citrus]

thank you, but could you please elaborate?

[dim shadow]

if you havent learned it you probably shouldnt be using it for the problems

cuz you wont be practicing solving limits the way they want you to

[inner pilot]

this is by using L-hospital law

[frozen citrus]

is there another way to solve this without the l'hospital?

[dim shadow]

u can probably simplify them with trig identities

[inner pilot]

I didn't use L-H here

first I think you know cos(2x)=cos^2 x-sin^2 x

then 1-cos^2 x=sin^2 x

therefore it's 2sin^2 x/5x^2

and sin x/ x=1 therefore sin^2 x/ x^2=1

therefore the ans is 2/5

[frozen citrus]

oh okayyy

how abt the 3?

[inner pilot]

My handwriting isn't good

[frozen citrus]

question, how is sin²x equal to 1-cosx?

[inner pilot]

sin^2 x=1-cos^2 x

and by using Square difference formula

[frozen citrus]

oh okayyy

[inner pilot]

1-cos^2 x=(1-cos x)(1+cos x)

[frozen citrus]

thank you!

.close

[devout snow]

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[frozen citrus]

.reopen

[devout snow]

✅

[frozen citrus]

what did you do here?

[inner pilot]

1-(cos^2 x -sin^2 x)=1-cos^2 x+sin^2 x

1-cos^2 x=sin^2 x

therefore

1-(cos^2 x -sin^2 x)=1-cos^2 x+sin^2 x=sin^2 x+sin^2 x

[frozen citrus]

oh okayy thank you

[devout snow]

@frozen citrus Has your question been resolved?

[devout snow]

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[frank star]

2x^2-3xy

b) and c) are way off, supposed to be = 20 and 1

And d, I don’t even know where to start. I can’t use a calculator so how do I do the root of 2?

[waxen harness]

What is the question asking?

[frank star]

Sorry, it’s asking what is the value of 2x^2-3xy if (insert picture) values

[waxen harness]

Well, the square root would not be a problem as x is squared and also xy will make it 2.

[frank star]

Okay so square root of 2 + square root of 2 = 2?

I mean multiplied not +

Well that makes sense

[waxen harness]

Yes

[frank star]

Alright let me try d) with this fresh reminder lol. I still don’t know what I’m doing in b) or c) though

In hindsight d) was very easy, I just didn’t think of it.

But I still don’t know b) and c), what’s going wrong? They’re supposed to become 20 and 1 respectively but as you see in my picture they become something else entirely. <@&286206848099549185>

[devout snow]

@frank star Has your question been resolved?

[frank star]

Huh maybe everyone is asleep

[frank star]

Sorry I forgot to ping you after

[polar chasm]

In b) you substituted 4 for y, while y should be -2

The same with c

[waxen harness]

Yes, substitute correctly and all will be fine. Ok?

[frank star]

thank you I will try this

[polar chasm]

For c) y=-1/3 and for b y=-2

[frank star]

yeah i put in the old 4 somehow

Let me just check that I can do it

Yeah we got it, than you bois

Thank*

Sometimes my brain doesn’t understand maths randomly for a day or so, then returns back to normal.

.close

[devout snow]

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[rapid spire]

help

[rapid spire]

move 6 to the left

3x/x-2 -6 >0

then what

cross multiply?

[polar chasm]

If you mean mutiply both sides by (x-2) then yes :)

[rapid spire]

u mutiply x-2 with -6

thats it?

why dont u touch bottom?

[mellow panther]

,rotate

[rapid spire]

u just time -6 with x-2

why dont u cross multiply for this equation

[rapid spire]

and why do u do this

.close

[devout snow]

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[devout snow]

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[gloomy ingot]

how do i do this ?

.close

[devout snow]

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[woven radish]

Mehdi_Moulati

[gloomy ingot]

how to do II

i have already worked out sin b = 12/13

and i used the double angle formula, but ive got a wrong answer

¯_(ツ)_/¯

[devout snow]

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[gloomy ingot]

how to do II
i have already worked out sin b = 12/13
and i used the double angle formula, but ive got a wrong answer
¯_(ツ)_/¯

[wooden wraith]

Do you mean part iii?

[gloomy ingot]

yeah sorry